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vichka [17]
3 years ago
5

How many moles of gold, a U, are in 3.60×10^-5 G of gold?

Chemistry
1 answer:
ch4aika [34]3 years ago
8 0

Answer:

See Below.

Explanation:

This is a conversion problem:

Using the molar mass of Gold (given in a periodic table) which is 196.97g/mol

you have

(3.60*10^-5 g) *\frac{1moles}{196.97g}

you always arrange the equation in a way to cancel whatever you don't want (grams) and leave what you do want (moles). Here grams cancel (top and bottom), so you're left with:

\frac{(3.60*10^-5 g) *(1)}{196.97} = 1.83*10^-7 moles of Gold

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Question 1<br><br> Of the following which displays a positive energetic process?
TiliK225 [7]

We can describe a positive energetic process as any process which increases the internal energy of the system.

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5 0
2 years ago
In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t
Arada [10]

Answer:

        \large\boxed{\large\boxed{0.529M}}

Explanation:

Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.

Hence, the rate law is:

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Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

      [A]=[A]_0e^{-kt}

You know [A]₀, k, and t, thus you can calculate [A].

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7 0
3 years ago
If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?
Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

  • <u>For TRIS base:</u>

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Volume of solution = 60 mL

Putting values in above equation, we get:

0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol

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  • To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

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pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of TRIS acid = 8.3

[\text{TRIS acid}]=\frac{0.005}{0.11}

[\text{TRIS base}]=\frac{0.012}{0.11}

pH = ?

Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

Hence, the pH of resulting solution is 8.7

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